2
$\begingroup$

I have a question that asks to calculate the following integral:

$$ \int_0^\infty {\frac{w\cdot \sin w}{4a^2+w^2}dw} $$

In the official solution they used partial fraction decomposition in order to later use Plancherel's identity:

$$ \frac{w\cdot \sin w}{4a^2+w^2} = $$ $$ \frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} = $$ $$ \frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2}) = $$ $$ \frac{\sin w}{w} - \frac{\sin w}{w} \cdot \frac{4a^2}{w^2+4a^2} $$

And then used Plancherel's identity.

But I didn't understand how to expand to partial fraction and in particular I didn't understand this equation: $$ \frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} = \frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2}) $$

Can you please explain how to expand the integrand into partial fraction?

$\endgroup$
  • $\begingroup$ $$\frac{\sin w}{w}\Big(1-\frac{4a^2}{w^2+4a^2}\Big)= \frac{\sin w}{w}\Big(\frac{w^2+4a^2}{w^2+4a^2}-\frac{4a^2}{w^2+4a^2}\Big)=\frac{\sin w}{w}\Big(\frac{w^2}{w^2+4a^2}\Big)$$ $\endgroup$ – Axion004 Sep 15 at 1:46
  • $\begingroup$ It is strange that the accepted answer is completely nonresponsive to the question asked: "Can you please explain how to expand the integrand into partial fraction?" $\endgroup$ – Eric Towers Sep 15 at 14:28
2
$\begingroup$

Consider the function:

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w},\space a>0$$

Rewrite the integrand as follows:

$$\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{w}{w}\cdot\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2+4\cdot a^2-4\cdot a^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\left(\frac{w^2+4\cdot a^2}{w^2+4\cdot a^2}-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}\cdot\left(1-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}-\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}$$

Then:

$$I(a)=\int_0^\infty {\frac{\sin (w)}{w}\text{d}w}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$

The left-hand integral is known as a Dirichlet integral and it can be derived that it evaluates to $\frac{\pi}{2}$:

$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$

Let $w\mapsto 2\cdot a\cdot w$:

$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{2\cdot a\cdot w}\cdot\frac{4\cdot a^2}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}$$

Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $a$ under the integral sign:

$$I'(a)=\frac{\text{d}}{\text{d}w}\left[\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}\right]=-\int_0^\infty {\frac{\partial}{\partial a}\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{w}{w}\cdot\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $a$ under the integral sign:

$$I''(a)=-2\cdot\frac{\text{d}}{\text{d}a}\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\partial}{\partial a}\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=4\cdot\int_0^\infty {\frac{w\cdot\sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Consider the original expression for $I(a)$:

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}$$

Let $w\mapsto 2\cdot a\cdot w$:

$$I(a)=\int_0^\infty {\frac{2\cdot a\cdot w\cdot \sin(2\cdot a\cdot w)}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\int_0^\infty {\frac{w\cdot \sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Recognize that

$$I''(a)=4\cdot I(a)\Rightarrow I''(a)-4\cdot I(a)=0$$

Solving the differential equation yields

$$I(a) = \text{c}_{1}\cdot e^{2\cdot a} + \text{c}_{2}\cdot e^{-2\cdot a}$$

Differentiate with respect to $a$ on both sides:

$$I'(a) = 2\cdot\left(\text{c}_{1}\cdot e^{2\cdot a} - \text{c}_{2}\cdot e^{-2\cdot a}\right)$$

According to the closed form of $I(a)$, as $a$ approaches $0$, $I(a\rightarrow 0)=\text{c}_{1}+\text{c}_{2}$.

According to the integral form of $I(a)$, as $a$ approaches $0$, $I(a\rightarrow 0)=\frac{\pi}{2}-\int_0^\infty {0\space\text{d}w}=\frac{\pi}{2}$.

According to the closed form of $I'(a)$, as $a$ approaches $0$, $I'(a\rightarrow 0)=2\cdot(\text{c}_{1}-\text{c}_{2})$.

According to the integral form of $I'(a)$, as $a$ approaches $0$, $I'(a\rightarrow 0)=-2\cdot\int_0^\infty {\frac{1}{w^2+1}\text{d}w}=-2\cdot\frac{\pi}{2}=-\pi$.

It can be derived that $\text{c}_{1}=0$ and $\text{c}_{2}=\frac{\pi}{2}$.

Then,

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}=\frac{\pi}{2}\cdot e^{-2\cdot a},\space a>0$$

$\endgroup$
2
$\begingroup$

There is no (nontrivial) partial fraction decomposition in the manipulations you described. The denominator is irreducible (over the reals), so that method normally is thwarted (but see below)

Consider $$ \frac{a}{a+b} = \frac{a+b-b}{a+b} = 1- \frac{b}{a+b} \text{.} $$


There is the possibility of a partial fractions decomposition, PFD, over the complex numbers. To perform any PFD, you must be able to combine fractions under the integral to a single fraction with a factored denominator. Since $$ 4a^2 + w^2 = (2a+\mathrm{i}w)(2a-\mathrm{i}w) \text{,} $$ both of which are linear in the variable of integration, $w$, so we would look for $u$ and $v$ such that $$ \frac{u}{2a+\mathrm{i}w} + \frac{v}{2a-\mathrm{i}w} \text{.} $$ Multiplying out and equating likes, we have $u = \mathrm{i}/2$ and $v = -\mathrm{i}/2$, so $$ \frac{w}{4a^2 + w^2} = \frac{\mathrm{i}/2}{2a+\mathrm{i}w} - \frac{\mathrm{i}/2}{2a-\mathrm{i}w} \text{.} $$ Of course, everyone knows $$ \int_0^\infty \; \frac{\mathrm{i}/2}{2a+\mathrm{i}w} \,\mathrm{d}w = \frac{1}{4} ((\pi +2 \mathrm{i} \, \mathrm{Shi}(2 a)) \cosh (2 a)-2 \mathrm{i} \, \mathrm{Ci}(-2 \mathrm{i} a) \sinh (2 a)) \text{,} $$ where $\mathrm{Shi}$ is the hyperbolic sine integral, $\cosh$ is the hyperbolic cosine, $\mathrm{Ci}$ is the cosine integral, and sinh is the hyperbolic sine. Or, to summarize, this is not the way to go for this problem.

$\endgroup$
0
$\begingroup$

$$1=\frac{w^2}{w^2+4a^2}+\frac{4a^2}{w^2+4a^2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.